Answer:
141.78 ft
Explanation:
When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.
Calculating the acceleration using one of Newton's equations of motion:
Note: The negative sign denotes deceleration.
When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2
Hence, we can find the minimum stopping distance using:
The minimum stopping distance is 141.78 ft.
Answer:
Magnitude of force on wheel B is 4 N
Explanation:
Given that
For wheel A
m= 1 kg
d= 1 m,r= 0.5 m
F=1 N
We know that
T= F x r
T=1 x 0.5 N.m
T= 0.5 N.m
T= I α
Where I is the moment of inertia and α is the angular acceleration
T= I α
0.5= 0.25 α
For Wheel B
m= 1 kg
d= 2 m,r=1 m
Given that angular acceleration is same for both the wheel
T= I α
T= 1 x 2
T= 2 N.m
Lets force on wheel is F then
T = F x r
2 = F x 1
So F= 2 N
Magnitude of force on wheel B is 2 N
Answer:
Change in momentum will be -4.4 kgm/sec
So option (A) is correct option
Explanation:
Mass of the ball is given m = 0.10 kg
Initial velocity of ball
And velocity after rebound
We have to find the change in momentum
So change in momentum is equal to ( here negative sign shows only direction )
So option (A) will be correct answer